A wire of resistance $$R$$ is connected in series with an inductor of reactance $${\omega L}.$$  Then quality factor of $$RL$$  circuit is

Solution :
We define the quality factor of the circuit as follows
Quality factor, $$Q = 2\pi \times \frac{{{\text{total energy stored in the circuit}}}}{{{\text{loss in energy in each cycle}}}}$$
But the total energy stored in circuit $$ = Li_{rms}^2$$
and the energy loss per second $$ = i_{rms}^2R$$
So, loss in energy per cycle $$ = \frac{{i_{rms}^2R}}{f}$$
Hence, quality factor, $$Q = 2\pi \times \frac{{Li_{rms}^2}}{{\frac{{i_{rms}^2R}}{f}}}$$
$$ = \frac{{2\pi fL}}{R} = \frac{{\omega L}}{R}$$